You have found the following ages (in years) of 6 gorillas. Those gorillas were randomly selected from the 21 gorillas at your local zoo: $ 20,\enspace 13,\enspace 8,\enspace 12,\enspace 6,\enspace 3$ Based on your sample, what is the average age of the gorillas? What is the standard deviation? You may round your answers to the nearest tenth.
Answer: Because we only have data for a small sample of the 21 gorillas, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $6$ samples and divide by $6$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\overline{x}} = \dfrac{20 + 13 + 8 + 12 + 6 + 3}{{6}} = {10.3\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {94.09} + {7.29} + {5.29} + {2.89} + {18.49} + {53.29}} {{6 - 1}} $ {s^2} = \dfrac{{181.34}}{{5}} = {36.27\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{36.27\text{ years}^2}} = {6\text{ years}} $ We can estimate that the average gorilla at the zoo is 10.3 years old. There is also a standard deviation of 6 years.